Hi all,
why printf("....%d" ,sizeof((int)(d ouble)(char) i)) always gives the
size of int ???
is it because sizeof doesn't evaluate its operand....??? 5 3605 aa*****@gmail.c om wrote:
Hi all,
why printf("....%d" ,sizeof((int)(d ouble)(char) i)) always gives the
size of int ???
is it because sizeof doesn't evaluate its operand....???
It is because your code is just the longer version of
printf("....%d" ,sizeof(int))
Bye, Jojo aa*****@gmail.c om said:
Hi all,
why printf("....%d" ,sizeof((int)(d ouble)(char) i)) always gives the
size of int ???
What did you want it to give the size of?
--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
On Apr 15, 1:47 pm, "Joachim Schmitz" <nospam.j...@sc hmitz-digital.de>
wrote:
aark...@gmail.c om wrote:
Hi all,
why printf("....%d" ,sizeof((int)(d ouble)(char) i)) always gives the
size of int ???
is it because sizeof doesn't evaluate its operand....???
It is because your code is just the longer version of
printf("....%d" ,sizeof(int))
Bye, Jojo
By doing printf("....%d" ,sizeof((int)(d ouble)(char) i)), you're
finally casting i to an int, so it would print the sizeof(int) value.
Correct me if I'm wrong
aark...@gmail.c om wrote:
why printf("....%d" ,sizeof((int)(d ouble)(char) i)) always gives
the size of int ???
is it because sizeof doesn't evaluate its operand....???
With the exception of variable length arrays in C99, yes.
The object i doesn't even have to be initialised. Even the
following is valid...
sizeof( (int) (int (*)(void)) -1 )
....even though there is no well defined conversion involved
at any stage.
--
Peter
Ivar wrote:
On Apr 15, 1:47 pm, "Joachim Schmitz" <nospam.j...@sc hmitz-digital.de>
wrote:
>aark...@gmail. com wrote:
>>Hi all,
>>why printf("....%d" ,sizeof((int)(d ouble)(char) i)) always gives the size of int ???
>>is it because sizeof doesn't evaluate its operand....???
It is because your code is just the longer version of printf("....%d ",sizeof(in t))
Bye, Jojo
By doing printf("....%d" ,sizeof((int)(d ouble)(char) i)), you're
finally casting i to an int, so it would print the sizeof(int) value.
Exactly, and that's what I said...
Correct me if I'm wrong
No need 8-)
Bye, Jojo This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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